Answer:
(a) 0.09
(b) 0.06
(c) 0.0018
(d) 0.9118
(e) 0.03
Step-by-step explanation:
Let <em>A</em> = boards have solder defects and <em>B</em> = boards have surface defects.
The proportion of boards having solder defects is, P (A) = 0.06.
The proportion of boards having surface-finish defects is, P (B) = 0.03.
It is provided that the events A and B are independent, i.e.
![P(A\cap B)=P(A)\times P(B)](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3DP%28A%29%5Ctimes%20P%28B%29)
(a)
Compute the probability that either a solder defect or a surface-finish defect or both are found as follows:
= P (A or B) + P (A and B)
![=P(A)+P(B)-P(A\cap B)+P(A\cap B)\\=P(A)+P(B)\\=0.06+0.03\\=0.09](https://tex.z-dn.net/?f=%3DP%28A%29%2BP%28B%29-P%28A%5Ccap%20B%29%2BP%28A%5Ccap%20B%29%5C%5C%3DP%28A%29%2BP%28B%29%5C%5C%3D0.06%2B0.03%5C%5C%3D0.09)
Thus, the probability that either a solder defect or a surface-finish defect or both are found is 0.09.
(b)
The probability that a solder defect is found is 0.06.
(c)
The probability that both defect are found is:
![P(A\cap B)=P(A)\times P(B)\\=0.06\times0.03\\=0.0018](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3DP%28A%29%5Ctimes%20P%28B%29%5C%5C%3D0.06%5Ctimes0.03%5C%5C%3D0.0018)
Thus, the probability that both defect are found is 0.0018.
(d)
The probability that none of the defect is found is:
![P(A^{c}\cup B^{c})=1-P(A\cup B)\\=1-P(A)-P(B)+P(A\cap B)\\=1-P(A)-P(B)+[P(A)\times P(B)]\\=1-0.06-0.03+(0.06\times0.03)\\=0.9118](https://tex.z-dn.net/?f=P%28A%5E%7Bc%7D%5Ccup%20B%5E%7Bc%7D%29%3D1-P%28A%5Ccup%20B%29%5C%5C%3D1-P%28A%29-P%28B%29%2BP%28A%5Ccap%20B%29%5C%5C%3D1-P%28A%29-P%28B%29%2B%5BP%28A%29%5Ctimes%20P%28B%29%5D%5C%5C%3D1-0.06-0.03%2B%280.06%5Ctimes0.03%29%5C%5C%3D0.9118)
Thus, the probability that none of the defect is found is 0.9118.
(e)
The probability that the defect found is a surface finish is 0.03.